In the figure, $A P = A Q$ and $B P = B Q .$

Prove that $A B$ is the bisector of $∠ P A Q$ and $∠ P B Q .$

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Solution

In $△ A P B$ and $△ A Q B$

$A P = A Q$ (Given)

$B P = B Q$ (Given)

$A B = A B$ (Common Side)

$∴ △ A P B ≅ △ A Q B$ (S.S.S. congruency)

$\Rightarrow ∠ P A B = ∠ Q A B$ (CPCT)

$\Rightarrow ∠ P B A = ∠ Q B A$ (CPCT)

Thus, AB is the bisector of $∠ P A Q$ and $∠ P B Q .$

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